idk how to solve this I am suppose to get type close to each other

Solve for x over the real numbers:
850 e^(0.055 x) = 195 e^(0.075 x)

850 e^(0.055 x) = 850 e^(11 x/200) and 195 e^(0.075 x) = 195 e^(3 x/40):
850 e^((11 x)/200) = 195 e^((3 x)/40)

Take the natural logarithm of both sides and use the identities log(a b) = log(a) + log(b) and log(a^b) = b log(a):
(11 x)/200 + log(850) = (3 x)/40 + log(195)

Subtract (3 x)/40 + log(850) from both sides:
-x/50 = log(195) - log(850)

Multiply both sides by -50:
Answer: | x = 50 log(850) - 50 log(195)= 73.61183954603077.......

850e^(0.055x)=195e^(0.075x)

$\begin{array}{|rcll|} \hline 850\cdot e^{0.055x} &=& 195\cdot e^{0.075x} \quad & | \quad : 195 \\ \frac{850}{195} \cdot e^{0.055x} &=& e^{0.075x} \quad & | \quad \cdot e^{-0.055x} \\ \frac{850}{195} &=& e^{0.075x} \cdot e^{-0.055x} \\ \frac{850}{195} &=& e^{0.075x-0.055x} \\ \frac{850}{195} &=& e^{0.02x} \\ \frac{170}{39} &=& e^{0.02x} \quad & | \quad \text{ln of both sides } \\ \ln(\frac{170}{39}) &=& \ln(e^{0.02x}) \\ \ln(\frac{170}{39}) &=& 0.02x\cdot \ln(e) \quad & | \quad \ln(e) = 1 \\ \ln(\frac{170}{39}) &=& 0.02x \quad & | \quad : 0.02 \\ \frac{ \ln(\frac{170}{39}) } {0.02} &=& x \\ \frac{ \ln(4.35897435897) } {0.02} &=& x \\ \frac{ 1.47223679092 } {0.02} &=& x \\ 73.6118395460 &=& x \\ \hline \end{array}$